## iop360 3 years ago Find all distinct roots(real or complex) of z^2 + (9+3i)z + (24+11i) in the form a+bi. Hint: You may have to complete the square

1. iop360

http://answers.yahoo.com/question/index?qid=20121124223053AAchfPc the bottom explanation of this link might be it, but i dont understand it

2. iop360

have you figured anything out?

3. sirm3d

$\large z^2+(9+3i)z+(24-11i)=0$$\large z^2+(9+3i)z=-(24-11i)$to complete squares, add (b/2)^2 to both sides, where b=9+3i $\large z^2+(9+3i)z+\left( \frac{ 9+3i }{ 2 } \right)^2=-(24-11i)+\left( \frac{ 9+3i }{ 2 } \right)^2$

4. iop360

yeah im at this step... what do you do next?

5. iop360

$(z + (\frac{ 9 + 3i }{ 2 }))^2 = -24 + 11i + 18 +\frac{ 27i }{ 2 }$

6. iop360

$z + (\frac{ 9+3i }{ 2 }) = \pm \sqrt{\frac{ -12 + 5i}{ 2 }}$

7. iop360

i dunno, lol

8. iop360

i dont get how the person completed the square in the yahoo link..

9. iop360

$z = -(\frac{ 9+3i }{ 2}) \pm \sqrt{\frac{ -12 + 5i }{ 2 }}$

10. iop360

it would be so much easier if the square root wasnt there...we could just add them up easily

11. iop360

from the yahoo answers link: z² + (2−4i)z = (11+10i) z² + (2−4i)z + (-3 -4i) = (11+10i) + (-3 -4i)

12. iop360

what i dont get is where the (-3-4i) comes from

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